For a recent project of mine, I wanted an adjustable resistance between 10Ω and 100kΩ, however none of the potentiometers I had went above 10kΩ. While I would have been able to put a few of them in series, that would be suboptimal as I would need to adjust all the potentiometers simultaneously.

I've tried connecting the potentiometer in series, parallel with resistors of other values, and different combinations of the two, but it seems like the "range" of the equivalent resistor (the difference between the maximum resistance and the minimum). It makes sense, as resistors in series increases the minimum and maximum the same amount, and parallel resistors (I think) always decrease the range, due to the formula R1R2/(R1+R2) being concave (negative second derivative).

In class, we've shown that using a transformer, one can transform a resistor of value R to (N1/N2)^2 R. However, my circuit is using DC, which makes the transformer method impossible.

Is there any way to transform a potentiometer with range [Rm, RM] to a potentiometer with range [Rm', RM'] where RM'-Rm' > RM-Rm in a DC-circuit?

  • 2
    \$\begingroup\$ General and quick answer: No there is no way to do what you ask. More detailed answer: It depends on what you want to do with that 100 kohm potmeter. If it is something that can only be done with a real pot., the answer remains no. If it is something that can be done using a circuit that behaves (under specific circumstances!!!) like a variable resistor then maybe yes. So explain what you're going to do with the 100 k ohm potmeter. \$\endgroup\$ Commented Dec 18, 2020 at 9:32
  • \$\begingroup\$ If you had ten in series, you wouldn't have to adjust them simultaneously. In effect, you would have a 10-turn potentiometer, just with a large panel area and a bit of inconvenience. \$\endgroup\$ Commented Dec 18, 2020 at 9:52
  • \$\begingroup\$ @Bimpelrekkie Hmm, I want to place the potentiometer between two legs of an IC, the discharge and tresh/trigger pin of a 555 timer to be specific. I was under the impression that all resistors were mostly equal, under what circumstances can I replace the potentiometer with a circuit that behaves like a variable resistor? Also, what would such a circuit be? \$\endgroup\$
    – xenia
    Commented Dec 18, 2020 at 9:56
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    \$\begingroup\$ If you want a bigger 555 time constant, use a bigger capacitor. \$\endgroup\$
    – Andy aka
    Commented Dec 18, 2020 at 10:06
  • \$\begingroup\$ I'll bet there IS a way to do it using an op-amp (or 2) and carefully applied positive feedback (yes positive)...and for sure you could do it (or most of it) with an OTA like the (ancient?) LM13600...but we are talking way way overkill at this point. (For example, you can easily use an op-amp to approximate an inductor using only Rs and Cs, etc.) Nat'l Semi / TI has an Application Note ti.com/lit/an/snla140d/snla140d.pdf that might be a starting point... \$\endgroup\$
    – Atomique
    Commented Dec 18, 2020 at 21:34

1 Answer 1


As I suspected, your question is a XY problem.

You don't need 100 k potmeter because there are other ways to get the same result using a 10 k potmeter.

You think that you need a 100 k potmeter for a 555 timer circuit. The timing of a 555 circuit is determined by a resistor (100k ohm) and a capacitor (of for example 100 nF).

That 100 kohm + 100 nF will give almost the same result if you use a 10 kohm resistor and a 1000 nF = 1 uF capacitor!

So just increase the value of the capacitor by a factor of 10. Also decrease all the resistors that are part of the 555's timing circuit by a factor of 10 (so a 100 kohm potmeter becomes a 10 kohm potmeter) and you're done.

Now I hope that you see, as I commented, that you have to explain what you're going to do with that potmeter. In case of a 555 timer circuit, the solution is simple and easy. Ideally you would include the schematic of what you're working on. When we would see that 555 timer, almost everyone here will know you only have to make the capacitor larger and then you can just use your 10 kohm potmeter.


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